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goldfiish [28.3K]
4 years ago
5

Find the distance for (3,5) and y=3x+2

Mathematics
1 answer:
Natali5045456 [20]4 years ago
5 0
There are standard formulas for this type of problem. However, it can also be solved by a combination of simple steps.

First, the shortest distance from the point (3,5) to the line y = x +4 (line1) will be along a straight line perpendicular to line1. Give the perpendicular line the name line2. Since the slope of line1 is 1, the slope of line2 will be -1.

Second, since line2 must go through (3,5) and also have a slope of -1, the point slope form can be used for line2:
y - 5 = (-1) (x -3)
So the equation of line2 is y = -x +8.

Third, the point of intersection of line1 and line2 can be found by solving the set of equations:
y = x +4
y = -x + 8
The solution of this set of two equations is x = 2, y = 6 i.e. the point (2,6) .

Fourth, the distance formula can be used to find the distance between (3,5) and (2,6)
d = sqrt( (3-2)2 + (5-6)2 ) = sqrt(2)
This is the desired distance.
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Answer:

1. {18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }

2a) md= 24 b) Q1=20.5 c) Q3= 30 3) Q3-Q1 =9.5 b) 19/48

Step-by-step explanation:

To answer this question the 1st and the 2nd we need to order the data entries. So from ordering from the lowest to the highest value:

1. {18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }

2. There are 25 entries.

{18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }

In odd quantities of observations, the Median equally separates it two parts.

md=24

b) To find out the 1st quartile, we can use this way:

Q_{1}=\frac{i}{4}(n+1)\\ Q_{1}=\frac{1}{4}(25+1)\\ Q_{1}=\frac{1}{4}(26)=6.5

Then 6.5 is between the 6th and 7th position. Let's find the mean of them, now:  

Q_1=\frac{20+21}{2}= 20.5

c) Similarly toThe Third Quartile or Upper Quartile

Q_{3}=\frac{i}{4}(n+1)\\ Q_{3}=\frac{3}{4}(25+1)\\ Q_{1}=\frac{3}{4}(26)=19.5

The 19th position and 20th position average is:\frac{30+30}{2} =30

3)

a) To find the Interquartile Range, we just need to find out the difference of the upper quartile and the lower one:[tex](Q_3-Q_1)Q_3-Q_1[/tex]

(30-20.5)=9.5

b) Interquartile Ratio is given by the quotient of the Interquartile Range over the Median

\frac{IQR}{md}=\frac{9.5}{24}=\frac{19}{48}

4) Since the  Relative  frequency Histogram asked is a one with 7 classes. Let's calculate how many  values.

k=1+3.32logn

7=1+3.32logn

6=3.32logn

n≈66

Each class must have an interval of 10 ages, for (91-18)/7≈ 10. Notice the orange line intercepts the midpoint of each interval.

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4 years ago
A student in an introductory statistics course investigated if there is evidence that the proportion of
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Complete question:

A student in an introductory statistics course investigated if there is evidence that the proportion of

milk chocolate M&M's that are green differs from the proportion of dark chocolate M&M's that are

green. She purchased a bag of each variety and her data are summarized in the following table:

Milk chocolate- Green:8 Not Green: 33 Total: 41

Dark Chocolate- Green: 4 Not green: 38 Total :42

a) Define the appropriate parameter(s) for this data.nd state the hypotheses for testing if the proportion of green M&M's differs for milk chocolate and dark chocolate M&M's.

b. The p-value is 0.32. Use this p-value to make a decision about these hypotheses with an alpha of 0.05. Be sure to word your decision in the context of the problem. Include an assessment of the strength of your evidence.

Step-by-step explanation:

a. P_{mc} = Proportion of milk chocolate M&M 's that are green

P_{dc} = Proportion of dark chocolate M&M 's that are green

H_{0} : P_{mc} = P_{dc}

H_{0} : P_{mc} ≠  P_{dc}

b. This p-value provides no evidence that the proportion of green candies differs for milk chocolate and dark chocolate M&M's

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Answer:

  1 : 12 : 4

Step-by-step explanation:

Let A, G, C represent the ages of Alex, George, and Carl, respectively. We are told that ...

  A = G +12

  C = 3A . . . . . . . we read this as "3 times as old"; not "3 times older"

  A + G + C = 68

Substituting for G and C, we have ...

  A + (A -12) +(3A) = 68

  5A = 80 . . . . . add 12

  A = 16 . . . . . . . divide by 5

The other ages are then ...

  G = A-12 = 16 -12 = 4

  C = 3A = 3·16 = 48

The ratios are ...

  G : C : A = 4 : 48 : 16

  G : C : A = 1 : 12 : 4

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3 years ago
A rental car company charges $40.18 per day to rent a car and $0.05 for every mile driven. Jacob wants to rent a car, knowing th
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Answer:

b) 270 ≥ 40.18d + 3.75

Step-by-step explanation:

Per day is a variable, so it's 40.18d

Per mile is also a variable so 0.05m

40.18d + 0.05m ≤ 270

40.18d + 0.05(75) ≤ 270

40.18d + 3.75 ≤ 270

5 0
3 years ago
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